Exam (elaborations) TEST BANK FOR Microeconomic Analysis 3rd Edition By Hal R. Varian (Solution Manual)

Exam (elaborations) TEST BANK FOR Microeconomic Analysis 3rd Edition By Hal R. Varian (Solution Manual) Answers to Exercises Microeconomic Analysis Third Edition Hal R. Varian University of California at Berkeley W. W. Norton & Company  New York  London ANSWERS Chapter 1. Technology 1.1 False. There are many counterexamples. Consider the technology generated by a production function f(x) = x2. The production set is Y = f(y;−x) : y  x2g which is certainly not convex, but the input requirement set is V (y) = fx : x p yg which is a convex set. 1.2 It doesn't change. 1.3 1 = a and 2 = b. 1.4 Let y(t) = f(tx). Then dy dt = Xn i=1 @f(x) @xi xi; so that 1 y dy dt = 1 f(x) Xn i=1 @f(x) @xi xi: 1.5 Substitute txi for i = 1;2 to get f(tx1; tx2) = [(tx1) + (tx2)] 1  = t[x 1 + x 2] 1  = tf(x1; x2): This implies that the CES function exhibits constant returns to scale and hence has an elasticity of scale of 1. 1.6 This is half true: if g0(x) 0, then the function must be strictly increasing, but the converse is not true. Consider, for example, the function g(x) = x3. This is strictly increasing, but g0(0) = 0. 1.7 Let f(x) = g(h(x)) and suppose that g(h(x)) = g(h(x0)). Since g is monotonic, it follows that h(x) = h(x0). Now g(h(tx)) = g(th(x)) and g(h(tx0)) = g(th(x0)) which gives us the required result. 1.8 A homothetic function can be written as g(h(x)) where h(x) is homogeneous of degree 1. Hence the TRS of a homothetic function has the 2 ANSWERS form g0(h(x)) @h @x1 g0(h(x)) @h @x2 = @h @x1 @h @x2 : That is, the TRS of a homothetic function is just the TRS of the underlying homogeneous function. But we already know that the TRS of a homogeneous function has the required property. 1.9 Note that we can write (a1 + a2) 1   a1 a1 + a2 x 1 + a2 a1 + a2 x 2 1  : Now simply dene b = a1=(a1 + a2) and A = (a1 +a2)1  . 1.10 To prove convexity, we must show that for all y and y0 in Y and 0  t  1, we must have ty + (1−t)y0 in Y . But divisibility implies that ty and (1 − t)y0 are in Y , and additivity implies that their sum is in Y . To show constant returns to scale, we must show that if y is in Y , and s 0, we must have sy in Y. Given any s 0, let n be a nonnegative integer such that n  s  n − 1. By additivity, ny is in Y ; since s=n  1, divisibility implies (s=n)ny = sy is in Y . 1.11.a This is closed and nonempty for all y 0 (if we allow inputs to be negative). The isoquants look just like the Leontief technology except we are measuring output in units of log y rather than y. Hence, the shape of the isoquants will be the same. It follows that the technology is monotonic and convex. 1.11.b This is nonempty but not closed. It is monotonic and convex. 1.11.c This is regular. The derivatives of f(x1; x2) are both positive so the technology is monotonic. For the isoquant to be convex to the origin, it is sucient (but not necessary) that the production function is concave. To check this, form a matrix using the second derivatives of the production function, and see if it is negative semidenite. The rst principal minor of the Hessian must have a negative determinant, and the second principal minor must have a nonnegative determinant. @2f(x) @x21 = −1 4x −32 1 x 12 2 @2f(x) @x1@x2 = 1 4x −1 2 1 x −12 2 @2f(x) @x22 = −1 4x 12 1 x −3 2 2 Ch. 2 PROFIT MAXIMIZATION 3 Hessian = " −1 4x −3=2 1 x1=2 2 1 4x −1=2 1 x −1=2 2 1 4x −1=2 1 x −1=2 2 −1 4x1=2 1 x −3=2 2 # D1 = −1 4x −3=2 1 x1=2 2 0 D2 = 1 16x −1 1 x −1 2 − 1 16x −1 1 x −1 2 = 0: So the input requirement set is convex. 1.11.d This is regular, monotonic, and convex. 1.11.e This is nonempty, but there is no way to produce any y 1. It is monotonic and weakly convex. 1.11.f This is regular. To check monotonicity, write down the production function f(x) = ax1 − p x1x2 + bx2 and compute @f(x) @x1 = a − 1 2x −1=2 1 x1=2 2 : This is positive only if a 1 2 q x2 x1 , thus the input requirement set is not always monotonic. Looking at the Hessian of f, its determinant is zero, and the determinant of the rst principal minor is positive. Therefore f is not concave. This alone is not sucient to show that the input requirement sets are not convex. But we can say even more: f is convex; therefore, all sets of the form fx1; x2: ax1 − p x1x2 + bx2  yg for all choices of y are convex. Except for the border points this is just the complement of the input requirement sets we are interested in (the inequality sign goes in the wrong direction). As complements of convex sets (such that the border line is not a straight line) our input requirement sets can therefore not be themselves convex. 1.11.g This function is the successive application of a linear and a Leontief function, so it has all of the properties possessed by these two types of functions, including being regular, monotonic, and convex. Chapter 2. Profit Maximization 4 ANSWERS 2.1 For prot maximization, the Kuhn-Tucker theorem requires the following three inequalities to hold  p @f(x) @xj −wj  x  j = 0; p @f(x) @xj − wj  0; x  j  0: Note that if x j 0, then we must have wj=p = @f(x)=@xj. 2.2 Suppose that x0 is a prot-maximizing bundle with positive prots (x0) 0. Since f(tx0) tf(x0); for t 1, we have (tx0) = pf(tx0) − twx0 t(pf(x0) − wx0) t(x0) (x0): Therefore, x0 could not possibly be a prot-maximizing bundle. 2.3 In the text the supply function and the factor demands were computed for this technology. Using those results, the prot function is given by (p;w) = p  w ap  a a−1 − w  w ap  1 a−1 : To prove homogeneity, note that (tp; tw) = tp  w ap  a a−1 − tw  w ap  1 a−1 = t(p;w); which implies that (p;w) is a homogeneous function of degree 1. Before computing the Hessian matrix, factor the prot function in the following way: (p;w) = p 1 1−aw a a−1  a a 1−a −a 1 1−a  = p 1 1−aw a a−1(a); where (a) is strictly positive for 0 a 1. The Hessian matrix can now be written as D2(p; !) = @2(p;w) @p2 @2(p;w) @p@w @2(p;w) @w@p @2(p;w) @w2 ! = 0 BBB@ a (1−a)2 p 2a−1 1−a w a a−1 − a (1−a)2 p a 1−a w 1 a−1 − a (1−a)2 p a 1−aw 1 a−1 a (1−a)2 p 1 1−a w 2−a a−1 1 CCCA (a): Ch. 2 PROFIT MAXIMIZATION 5 The principal minors of this matrix are a (1 − a)2 p 2a−1 1−a w a a−1 (a) 0 and 0. Therefore, the Hessian is a positive semidenite matrix, which implies that (p;w) is convex in (p;w). 2.4 By prot maximization, we have jTRSj = @f @x1 @f @x2 = w1 w2 : Now, note that ln(w2x2=w1x1) = −(ln(w1=w2) + ln(x1=x2)): Therefore, d ln(w2x2=w1x1) d ln(x1=x2) = d ln(w1=w2) d ln(x2=x1) − 1 = dln jTRSj d ln(x2=x1) − 1 = 1= − 1: 2.5 From the previous exercise, we know that ln(w2x2=w1x1) = ln(w2=w1) + ln(x2=x1); Dierentiating, we get d ln(w2x2=w1x1) d ln(w2=w1) = 1− dln(x2=x1) d ln jTRSj = 1−: 2.6 We know from the text that Y O  Y  Y I. Hence for any p, the maximum of py over YO must be larger than the maximum over Y , and this in turn must be larger than the maximum over Y I. 2.7.a We want to maximize 20x − x2 − wx. The rst-order condition is 20 − 2x − w = 0. 2.7.b For the optimal x to be zero, the derivative of prot with respect to x must be nonpositive at x = 0: 20−2x−w 0 when x = 0, or w  20. 2.7.c The optimal x will be 10 when w = 0. 2.7.d The factor demand function